智能指针是否违反了最小化头文件中#includes的原则？

Question

I prefer to minimize the use of #include in my header files, using forward declarations wherever possible, and I believe this is considered good practice.

It works great if I have a method declaration like:

bool IsFlagSet(MyObject *pObj);

However if I have typedef Ptr<MyObject> MyObjectPtr in MyObject.h and the API changes to:

bool IsFlagSet(MyObjectPtr pObj);

Do I not now have to #include "MyObject.h" ? Is there any way around this, or is it just the price one pays for using smart pointers?

Answer 1

No, you do not have to. You can define a type alias for an incomplete class, and template arguments can be incomplete types (see Paragraph 14.3.1/2 of the C++11 Standard):

#include <memory>

struct C;

typedef std::shared_ptr<C> ptrC; // C is incomplete here

struct C { void foo() { } };

int main()
{
    ptrC p = std::make_shared<C>();
    p->foo();
}

As correctly mentioned by Pubby in the comments, function declarations do not require the types mentioned in their signature to be complete either:

struct C;

void foo(C); // C is incomplete here

struct C { };

#include <iostream>

void foo(C)
{
    std::cout << "foo(C)" << std::endl;
}

int main()
{
    C c;
    foo(c);
}

Answer 2

No, std::shared_ptr<T> is explicitly designed to work when T is only forward-declared. Of course, this does not work for all cases, but the principle is the same as for a plain pointer. If T is forward-declared, you can do anything with std::shared_ptr<T> that you could do with T* .

Answer 3

you can use typedef with incomplete type.

But isn't it better to use full name of the smart pointer type?

MyClassPtr is of course much shorter, but std::unique_ptr<MyClass> actually tells us how to use this pointer. So for not very long names I suggest using full names of smart pointers.